Calculate the derivate $dx/dy$ using $\int_0^x \sqrt{6+5\cos t} \, dt + \int_0^y \sin t^2 \, dt = 0$ I want to calculate $\frac{dx}{dy}$ using the equation below.
$$\int_0^x \sqrt{6+5\cos t}\;dt + \int_0^y \sin t^2\;dt = 0$$
I don't even know from where to start. Well I think that I could first find the integrals and then try to find the derivative. The problem with this approach is that I cannot find the result of the first integral.
Can someone give me a hand here?
 A: For $\int_0^x f(t)\,dt + \int_0^y g(t)\,dt = 0$, I'd start by rewriting to
$$\int_0^x f(t)\,dt = -\int_0^y g(t)\,dt = u$$
where I've introduced a new variable for the common value of the two sides. This allows us to imagine that the curve is parameterized by $u$.
The fundamental theorem of calculus then gives us $\frac{du}{dx}=f(x)$ and $\frac{du}{dy}=-g(y)$, and we can then get $\frac{dx}{du}$ and $\frac{dy}{du}$ as the reciprocals of those, and finally we can find $\frac{dx}{dy}$ by implicit differentiation. Thus we don't actually need to evaluate the two integrals!
(This is exactly the reverse procedure from finding the curve by solving the resulting differential equation by separation of the variables).
A: HINT: You have $$f(x)=\int_0^x\sqrt{6+5\cos t}\,dt=-\int_0^y\sin t^2 \,dt=g(y)\;.$$ What are $\dfrac{df}{dx}$ and $\dfrac{dg}{dy}$ according to the fundamental theorem? And when you have $\dfrac{df}{dx}$, what can you multiply it by to get $\dfrac{df}{dy}$?
A: First, differentiate both sides with respect to $x$:
$$0=\frac{d}{dx} \left(\int_0^x \sqrt{6+5\cos t}\;dt + \int_0^y \sin t^2\;dt\right) = \sqrt{6+5\cos x} + (\sin y^2)\frac{dy}{dx}$$
Now we have a differential equation:
$$
\sqrt{6+5\cos x} + (\sin y^2)\frac{dy}{dx} = 0.
$$
Separate variables:
$$
(\sin y^2)\;dy = -\sqrt{6+5\cos x}\;dx
$$
Now the problem is to find two antiderivatives.
